Point \(D\) is on side \(QR\) of \(\triangle PQR\) with \(PD\perp QR\). Is \(\triangle PQD \sim \triangle RPD\)? Why?
No (not necessarily).
Step 1: We are given that \(PD\) is perpendicular to \(QR\). This means \(\angle QDP = 90^\circ\) and \(\angle RDP = 90^\circ\).
Step 2: For two triangles to be similar, we need either:
Step 3: In \(\triangle PQD\) and \(\triangle RPD\), we already have one common fact: both contain a right angle at point \(D\).
Step 4: But we also need at least one more equal angle (or a clear ratio of sides). For example, we would need \(\angle QPD = \angle RPD\). But this is not guaranteed just from the information given.
Step 5: Because we only know about one equal angle (the right angle), and we have no other angle or side ratio, we cannot apply the similarity rules.
Therefore, we cannot conclude that \(\triangle PQD\) is similar to \(\triangle RPD\).